For the function $f(x)=\sin x-\sqrt{3}\cos x-x$ on the interval $[0,\pi ]$, the stationary points will be:

Stationary points occur when $f^{\prime }(x)=0$.$f^{\prime }(x)=\cos x+\sqrt{3}\sin x-1=0$$\cos x+\sqrt{3}\sin x=1$ Divide by $2$:$\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x=\frac{1}{2}$$\cos \left(\frac{\pi }{3}\right)\cos x+\sin \left(\frac{\pi }{3}\right)\sin x=\frac{1}{2}$$\cos \left(x-\frac{\pi }{3}\right)=\cos \left(\frac{\pi }{3}\right)$ General solution: $x-\frac{\pi }{3}=2n\pi \pm \frac{\pi }{3}$.For $n=0$:$x-\frac{\pi }{3}=\frac{\pi }{3}⟹x=2\pi /3$$x-\frac{\pi }{3}=-\frac{\pi }{3}⟹x=0$ For $n=1$:$x-\frac{\pi }{3}=2\pi -\frac{\pi }{3}⟹x=2\pi -\frac{2\pi }{3}+\frac{\pi }{3}⟹x=2\pi$ (not in $[0,\pi ]$)In the interval $[0,\pi ]$, the stationary point is $x=2\pi /3$. (Note: $x=0$ is an endpoint).